In ZF you can't prove that there is a finitely additive measure on $\mathcal P(\mathbb N)$ (giving all singletons measure 0 and $\mathbb N$ measure 1). The existence of such a measure gives you a subset of $\mathbb R$ without the Baire property, and such a set cannot be constructed without some help of the axiom of choice. In other words, there is no explicit "construction" of a non-trivial finitely additive measure.

Since you don't like the ultrafilter measure, here is one that is less trivial:

Instead of $\mathbb N$, we use the integers $\mathbb Z$ as the underlying set. For $n\in\mathbb N$ let $d_n(A)=\frac{|A\cap\{-n,\dots,n\}|}{2n+1}$. The sequence $(d_n(A))_{n\in\mathbb N}$ typically does not converge. We fix this using an ultrafilter $U$ on $\mathbb N$ as follows:

For each $A\subseteq\mathbb Z$ let $\mu(A)$ be the unique element of the set $\bigcap_{S\in U}\mbox{cl}(\{d_n(A):n\in S\})$. Now $\mu$ is a finitely additive measure on $\mathcal P(\mathbb Z)$ that is even translation invariant and hence very different from the ultrafilter measure that you find too trivial.

Notice that I have just proved the wellknown fact that the integers are an amenable group.

In ZF you can't prove that there is a finitely additive measure on $\mathcal P(\mathbb N)$ (giving all singletons measure 0 and $\mathbb N$ measure 1). The existence of such a measure gives you a subset of $\mathbb R$ without the Baire property, and such a set cannot be constructed without some help of the axiom of choice. In other words, there is no explicit "construction" of a non-trivial finitely additive measure.

Since you don't like the ultrafilter measure, here is one that is less trivial:

Instead of $\mathbb N$, we use the integers $\mathbb Z$ as the underlying set. For $n\in\mathbb N$ let $d_n(A)=\frac{|A\cap\{-n,\dots,n\}|}{2n+1}$. The sequence $(d_n(A))_{n\in\mathbb N}$ typically does not converge. We fix this using an ultrafilter $U$ on $\mathbb N$ as follows:

For each $A\subseteq\mathbb Z$ let $\mu(A)$ be the unique element of the set $\bigcap_{S\in U}\mbox{cl}(\{d_n(A):n\in S\})$. Here the closure is taken in $\mathbb R$. Now $\mu$ is a finitely additive measure on $\mathcal P(\mathbb Z)$ that is even translation invariant and hence very different from the ultrafilter measure that you find too trivial.

Notice that I have just proved the well known fact that the integers are an amenable group.

In ZF you can't prove that there is a finitely additive measure on $\mathcal P(\mathbb N)$ (giving all singletons measure 0 and $\mathbb N$ measure 1). The existence of such a measure gives you a subset of $\mathbb R$ without the Baire property, and such a set cannot be construct without some instance of the axiom of choice. In other words, there is no explicit "construction" of a non-trivial finitely additive measure.

In ZF you can't prove that there is a finitely additive measure on $\mathcal P(\mathbb N)$ (giving all singletons measure 0 and $\mathbb N$ measure 1). The existence of such a measure gives you a subset of $\mathbb R$ without the Baire property, and such a set cannot be constructed without some help of the axiom of choice. In other words, there is no explicit "construction" of a non-trivial finitely additive measure.

Since you don't like the ultrafilter measure, here is one that is less trivial:

Instead of $\mathbb N$, we use the integers $\mathbb Z$ as the underlying set. For $n\in\mathbb N$ let $d_n(A)=\frac{|A\cap\{-n,\dots,n\}|}{2n+1}$. The sequence $(d_n(A))_{n\in\mathbb N}$ typically does not converge. We fix this using an ultrafilter $U$ on $\mathbb N$ as follows:

For each $A\subseteq\mathbb Z$ let $\mu(A)$ be the unique element of the set $\bigcap_{S\in U}\mbox{cl}(\{d_n(A):n\in S\})$. Now $\mu$ is a finitely additive measure on $\mathcal P(\mathbb Z)$ that is even translation invariant and hence very different from the ultrafilter measure that you find too trivial.

Notice that I have just proved the wellknown fact that the integers are an amenable group.